Take a breath before you start this one. “Atoms” looks intimidating because it arrives with three famous names, a radioactive source, five spectral series and a derivation that runs to half a page. It is none of those things underneath. This entire chapter is one picture and one equation. The picture is that an atom is almost entirely empty space with a ridiculously small, ridiculously heavy dot at the centre. The equation is Coulomb attraction supplying the centripetal force. Everything else in the chapter — the radius of the nth orbit, the velocity, the energy, the whole hydrogen spectrum — falls out of those two things.
These atoms class 12 physics notes teach the chapter from zero. No prior memory of Class 11 rotational mechanics is assumed; we will rebuild the one formula we need. You will get the full Bohr model of hydrogen atom derivation, the complete radius of nth orbit derivation class 12 examiners actually ask for, a hydrogen spectrum series formula list you can revise in ninety seconds, and a worksheet of atoms class 12 physics important questions with fully worked answers. Chapter 12 sits in Unit VIII, which carries 12 marks together with Chapter 13 (Nuclei) and Unit VII (Dual Nature) in the 70-mark Class 12 paper — small-looking, but it is some of the cheapest marks on the whole syllabus because the questions repeat.
One promise: by the end of this page you will be able to look at any wavelength a question throws at you and say, out loud and with confidence, which series it belongs to and which two levels made it. That single skill is worth more than memorising the formulas, and we build it deliberately in a later section.
Meet Your Tutor
Atomic Physics is a short chain of evidence: scattering reveals the nucleus, spectra reveal energy levels and Bohr’s model connects the two. I will help you keep assumptions, formulae and units together so each numerical tells the same physical story as the theory.
What You’ll Learn
- Start Here: The Scale Model of the Atom (Marble and Kilometre)
- Why Thomson’s Plum-Pudding Atom Had to Go
- Alpha-Particle Scattering Experiment (Geiger–Marsden): What Was Actually Done
- The Three Observations and What Each One Implies
- Rutherford’s Nuclear Model of the Atom
- Size of the Nucleus, Impact Parameter and Distance of Closest Approach
- Why Rutherford’s Model Fails: Instability and the Continuous Spectrum
- Bohr Model of Hydrogen Atom: The Three Postulates
- Angular Momentum Quantisation: What mvr = nh/2π Actually Says
- Radius of nth Orbit Derivation Class 12 (Full Working)
- Velocity of the Electron in the nth Orbit — and the c/137 Surprise
- Energy of the Electron in the nth Orbit: KE, PE and Total Energy
- Time Period and Orbital Frequency of the Electron
- The 2-Spine Trick: One Memory Device for r ∝ n², v ∝ 1/n, E ∝ 1/n², T ∝ n³
- Hydrogen Line Spectra (Qualitative Treatment Only) and the Energy-Level Diagram
- Hydrogen Spectrum Series Formula List: Lyman, Balmer, Paschen, Brackett, Pfund
- Which Series Does This Line Belong To? A Four-Step Decision Procedure
- Emission vs Absorption Spectra, Excitation Energy and Ionisation Energy
- Limitations of Bohr’s Model
- Reference Only — Adjacent Topics Not in Your 2026-27 CBSE Exam
Your Game Plan
- Read the scale model section first and do not skip it. It is thirty seconds long and it makes the rest of the chapter obvious instead of arbitrary.
- Learn the three observations of the α-scattering experiment and, for each one, the single sentence it forces on you. Examiners ask for the implication, not the apparatus.
- Write the radius derivation out on paper five times until you can do it without looking. Every other Bohr result is one line away from it.
- Memorise the 2-Spine Trick in this page instead of memorising four separate powers of n. You will stop mixing them up permanently.
- Drill the which-series decision procedure until it is automatic. Spectra questions are then free marks.
- Finish the ten-question worksheet at the bottom on paper. Reading a solution feels like learning; writing one is learning. More chapters are in the study notes library when you are done.
Study Notes
Start Here: The Scale Model of the Atom (Marble and Kilometre)
An atom is about 10−10 m across in radius — one ångström. The nucleus buried at its centre is about 10−15 m in radius — one fermi or femtometre. Those two numbers are written so close together on the page that your brain files them as “both small”. They are not both small. One is a hundred thousand times the other.
So here is the image I want you to carry through every single section below. Blow the nucleus up to the size of a 1 cm glass marble. On that scale, where is the edge of the atom? Multiply by 100 000: the atom’s radius becomes 1 cm × 105 = 100 000 cm = 1 kilometre. Put the marble on the pitch at your local cricket ground and the nearest electron is a kilometre away — past the boundary, past the gate, past the main road. In between: nothing. Not thin gas. Nothing.
That factor of 105 in radius becomes (105)3 = 1015 in volume. The nucleus occupies roughly one part in a thousand million million of the atom’s volume — and yet it carries about 99.95% of the atom’s mass (for hydrogen, the proton is 1836 times heavier than the electron, so the proton’s share is 1836/1837 ≈ 0.9995).
One step. Volume goes as radius cubed, so the fraction is (rnuc/ratom)3.
= (10−15/10−10)3 = (10−5)3 = 10−15.
One part in 1015. If the atom were a cube of side 1 metre, the nucleus would be a cube of side 10 micrometres — a speck of dust you would need a microscope to see. Hold on to how absurd that is; the whole α-scattering result depends on it.
Why Thomson’s Plum-Pudding Atom Had to Go
Before 1911 the best available picture was J. J. Thomson’s model, nicknamed the plum-pudding (or watermelon) model. The idea: an atom is a uniform sphere of positive charge, roughly 10−10 m in radius, with the electrons stuck inside it like raisins in a pudding or seeds in a watermelon. The positive charge is smeared out over the whole atom.
It was not a silly model. It explained two real facts. First, atoms are electrically neutral overall — the smeared positive charge exactly cancels the electrons. Second, electrons can be knocked out of atoms fairly easily, which fits loosely embedded raisins.
But look at what “smeared out” costs you. If you spread the whole positive charge of a gold atom over a sphere of radius 10−10 m, the electric field anywhere inside is feeble. A fast, heavy, doubly-charged projectile crossing such an atom would get nudged by a fraction of a degree, and crossing a thousand atoms in a row those tiny random nudges would add up like coin tosses — to maybe a degree or two in total. Thomson’s model makes a hard, testable prediction: large-angle deflection is impossible. Not unlikely. Impossible.
If you want to refresh why a diffuse charge distribution gives a weak field while a concentrated one gives an enormous field, the inverse-square machinery behind it is set out in Coulomb’s law and electric potential energy in the Electric Charges and Fields notes. Everything in this chapter is Coulomb’s law wearing different hats.
Alpha-Particle Scattering Experiment (Geiger–Marsden): What Was Actually Done
Ernest Rutherford handed his students Hans Geiger and Ernest Marsden what looked like a boring confirmation job: fire α-particles at a thin metal foil and measure the small deflections. The set-up, in the order the α-particle meets it:
- A source of α-particles of about 5.5 MeV, sitting inside a lead block. An α-particle is a helium nucleus: charge +2e, mass about 4 u — roughly 7300 times the mass of an electron. That mass ratio matters enormously, and we will come back to it.
- Lead slits that collimate the emitted particles into a narrow, well-defined beam. Without this you have a spray, not a beam, and you cannot measure an angle.
- A thin gold foil, about 2 × 10−7 m thick. Gold was chosen because it is highly malleable, so it can be beaten into a foil only a few hundred to a thousand atoms thick. Thin is essential: you want each α-particle to have at most one significant encounter, not a hundred.
- A rotatable detector — a zinc sulphide screen viewed through a microscope. Each α-particle striking the screen makes a tiny flash (a scintillation). Geiger and Marsden sat in the dark and counted flashes per minute at each scattering angle θ.
The scattering angle θ is measured from the original direction of the beam. θ = 0° means “kept going straight”; θ = 180° means “came straight back at me”.
Step 1 — convert to joules. K = 5.5 × 106 × 1.602 × 10−19 = 8.81 × 10−13 J.
Step 2 — use K = ½mv².
v = √(2K/m) = √(2 × 8.81 × 10−13 / 6.64 × 10−27) = √(2.653 × 1014) = 1.63 × 107 m s−1.
Comment: that is 5.4% of the speed of light — fast, but comfortably non-relativistic, so ½mv² is legitimate here. It is also why the α-particle is such a good probe: it is fast and heavy enough that only something extraordinarily strong can turn it around.
The Three Observations and What Each One Implies
This is the highest-value paragraph in the chapter. Learn the three observations paired with their implications. Examiners almost never ask for one without the other.
| Observation | What was seen | What it forces you to conclude |
|---|---|---|
| 1. Most pass through | The overwhelming majority of α-particles went through the foil almost undeviated (θ close to 0°) | The atom is mostly empty space. There is nothing along most straight lines through an atom to interact with. |
| 2. A few deflected a lot | A small fraction were deflected through large angles | The positive charge is concentrated in a tiny volume, not smeared out. Only a very intense, very localised field can bend a 5.5 MeV projectile sharply. |
| 3. Very few bounced back | About 1 in 8000 was turned through more than 90°; a handful came almost straight back | That concentrated charge is also extremely massive — far heavier than the α-particle. A light target cannot reverse a heavy projectile. |
Observation 3 deserves Rutherford’s own reaction, because it captures the physics exactly: he described it as being about as credible as firing a shell at tissue paper and having it come back and hit you. The point of the image is a mass argument. Roll a cricket ball at a row of ping-pong balls and it never comes back — it cannot, because momentum conservation forbids a heavy object bouncing backwards off a much lighter one. Roll it at a brick wall and it does. So whatever the α-particle met inside a gold atom behaved like a brick wall, not a ping-pong ball.
A few strongly deflected → positive charge is concentrated in a tiny central region.
A very few back-scattered → that region holds almost all the mass.
Three observations, three implications, and together they define the nucleus.
Notice how the scale model predicts the numbers too, not just the words. Fire bullets at random at a 1 km-radius sphere containing one marble at its centre. Almost every bullet misses the marble by an enormous margin — observation 1. A tiny fraction skims close enough to be swung hard — observation 2. An almost vanishing fraction happens to be aimed nearly dead at the marble and is thrown straight back — observation 3. Same experiment, told with a marble.
Rutherford’s Nuclear Model of the Atom
In 1911 Rutherford put the three implications together into a model. Learn it as four statements:
- All the positive charge and nearly all the mass of an atom are concentrated in an extremely small central region called the nucleus, of radius of order 10−15 m.
- The nucleus is surrounded by electrons which occupy the remaining volume — a region about 105 times larger in radius.
- The electrons revolve around the nucleus in circular orbits, exactly the way planets orbit the Sun. For this reason the model is often called the planetary or nuclear model.
- The electrostatic (Coulomb) attraction between the nucleus and an electron supplies the necessary centripetal force. The atom as a whole is electrically neutral, so the number of orbital electrons equals the number of protons, Z.
Statement 4 is the equation that runs the rest of the chapter, so write it now and keep it visible:
(1/4πε0) · Ze²/r² = mv²/r
Left side: Coulomb attraction between a nucleus of charge +Ze and an electron of charge −e, separated by r. Right side: the centripetal force needed to keep mass m moving in a circle of radius r at speed v. For a hydrogen atom, Z = 1 and the equation becomes (1/4πε0)e²/r² = mv²/r.
A neat by-product: multiply both sides by r and you get (1/4πε0)Ze²/r = mv², so ½mv² = (1/4πε0)Ze²/2r. That is the kinetic energy, free of charge, before we have even mentioned Bohr. Keep it — we use it in the energy section.
Size of the Nucleus, Impact Parameter and Distance of Closest Approach
Three quantities live in this section and students routinely confuse them. Separate them clearly right now.
| Quantity | Symbol | What it actually is |
|---|---|---|
| Impact parameter | b | The perpendicular distance from the nucleus to the original straight-line path of the α-particle, if the α-particle had not been deflected. It is the “aim”. Small b = well aimed = big deflection. |
| Distance of closest approach | r0 | How close the α-particle actually gets to the nucleus before the repulsion turns it around. Smallest for a head-on collision (b = 0, θ = 180°). |
| Nuclear radius | R | The actual size of the nucleus. The head-on r0 gives an upper limit on R: the nucleus can be no bigger than the closest the α-particle got. |
Impact parameter formula. Rutherford’s scattering analysis gives, for an α-particle of charge 2e and kinetic energy K = ½mv² scattered by a nucleus of atomic number Z:
b = Ze² cot(θ/2) / (4πε0 · ½mv²)
You are not asked to derive this. You are asked to read it. Notice the cotangent. As θ → 180°, cot(θ/2) → cot 90° = 0, so b → 0: back-scattering happens only for a nearly head-on hit. As θ → 0°, cot(θ/2) → ∞, so b is huge: a distant miss barely deflects at all. That single line is the mathematical version of the three observations.
Distance of closest approach. This one you should be able to derive in two lines, because it is a pure energy conservation argument. Fire an α-particle of kinetic energy K dead-on at a nucleus of charge +Ze. As it approaches, the repulsive Coulomb potential energy climbs and the kinetic energy falls. At the closest point it has momentarily stopped: all the kinetic energy has become electrostatic potential energy.
K = (1/4πε0) · (2e)(Ze)/r0 ⇒ r0 = (1/4πε0) · 2Ze²/K
The factor 2Ze² is (2e) × (Ze) — the two charges multiplied. Students lose the 2 constantly. The α-particle carries two elementary charges; do not let that 2 vanish.
Step 1 — energy in joules. K = 5.5 × 106 × 1.602 × 10−19 = 8.812 × 10−13 J.
Step 2 — substitute.
r0 = 8.99 × 109 × 2 × 79 × (1.602 × 10−19)² / 8.812 × 10−13
Numerator = 8.99 × 109 × 158 × 2.567 × 10−38 = 3.646 × 10−26
r0 = 3.646 × 10−26 / 8.812 × 10−13 = 4.14 × 10−14 m = 41.4 fm.
Why it works — and what it tells you. 41.4 fm is an upper limit on the gold nuclear radius, and it is about 2500 times smaller than the atom (10−10 m). A 5.5 MeV α-particle is simply not energetic enough to reach the nuclear surface, which sits at around 7 fm — it is stopped by pure electrostatic repulsion well before it gets there.
Step 1. cot(θ/2) = cot 45° = 1, so b = Ze²/(4πε0K).
Step 2. b = 8.99 × 109 × 79 × 2.567 × 10−38 / 8.812 × 10−13 = 1.823 × 10−26 / 8.812 × 10−13 = 2.07 × 10−14 m = 20.7 fm.
Sanity check: compare with the head-on answer above. r0 (head-on) = 41.4 fm and b (for 90°) = 20.7 fm = exactly half of it. That is not a coincidence — at θ = 90° the formula reduces to half the head-on distance, which is a lovely self-check to remember.
(a) Rearrange r0 = 2Ze²/(4πε0K) for K:
K = 2Ze²/(4πε0r0) = 8.99 × 109 × 158 × 2.567 × 10−38 / 7.0 × 10−15
= 3.646 × 10−26 / 7.0 × 10−15 = 5.21 × 10−12 J
In MeV: 5.21 × 10−12 / (1.602 × 10−13) = 32.5 MeV.
(b) That is about six times the energy of the α-particles Geiger and Marsden actually had. So their experiment never touched the nucleus at all — it only ever probed the pure Coulomb field outside it. That is precisely why the result was so clean: with no nuclear force in play, the inverse-square law held perfectly, and the measured angular distribution matched Rutherford’s prediction exactly. Historically, discrepancies at much higher energies were later the first hint of the nuclear force.
Why Rutherford’s Model Fails: Instability and the Continuous Spectrum
Rutherford’s model got the structure right and the dynamics catastrophically wrong. Two failures, and you need both.
Failure 1: the atom should collapse. An electron moving in a circle is accelerating — its direction is constantly changing, so it has centripetal acceleration even at constant speed. Classical electromagnetism is unambiguous on this point: an accelerating charge radiates electromagnetic energy. Radiating means losing energy. Losing energy means the orbit must shrink. So the electron should spiral inwards and crash into the nucleus.
How fast? A classical calculation gives a collapse time of the order of 10−8 s — ten nanoseconds. Hydrogen atoms have been sitting around stably for about 13.8 billion years. The model is not slightly wrong; it is wrong by roughly 25 orders of magnitude. Rutherford’s atom cannot explain the stability of the atom.
Failure 2: the spectrum should be continuous. As the electron spirals in, its orbital radius decreases smoothly and so its orbital frequency increases smoothly. The frequency of the radiation it emits equals its orbital frequency, so the emitted frequency should also sweep up smoothly. The result would be a continuous spectrum — an unbroken smear of all frequencies, like a rainbow with no gaps.
What is actually observed when you pass a discharge through hydrogen gas and send the light through a prism? A handful of sharp, discrete, brightly coloured lines at very specific wavelengths, with darkness in between. Not a smear. A barcode. Rutherford’s model has no mechanism whatsoever for producing discrete lines.
(ii) Spectrum: a continuously shrinking orbit gives a continuously changing frequency, so the model predicts a continuous emission spectrum. The observed atomic spectrum is a set of discrete lines.
Both failures have the same root cause: Rutherford used classical physics where classical physics does not apply. Bohr’s fix was not to find a better classical mechanism. It was to forbid the classical behaviour by decree, and see what followed. That is the next section, and it is the heart of the chapter.
Bohr Model of Hydrogen Atom: The Three Postulates
In 1913 Niels Bohr kept Rutherford’s nucleus, kept the circular orbits, and then bolted on three postulates that classical physics has no right to allow. He did not derive them. He assumed them, because assuming them produced the hydrogen spectrum to four decimal places. That is how physics sometimes works.
Here are the three postulates. Learn them as orbits → angular momentum → jumps.
What it buys you: the stability of the atom. The electron is simply not permitted to radiate while it stays in an allowed orbit.
L = mvnrn = nh/2π, n = 1, 2, 3, …where h is Planck’s constant and n is the principal quantum number.
What it buys you: only certain radii and certain energies are allowed. This is the postulate that turns a continuum into a ladder.
hν = Ei − EfWhat it buys you: discrete spectral lines. Discrete energy differences give discrete photon energies give discrete wavelengths. The barcode is explained.
Note carefully what the third postulate does not say. It does not say the emitted frequency equals the orbital frequency of the electron. That was Rutherford’s (classical) assumption and it was wrong. Bohr severs the link entirely: the photon frequency is set by the difference between two energy levels, and has nothing to do with how fast the electron is going round.
Angular Momentum Quantisation: What mvr = nh/2π Actually Says
Postulate 2 is where most students nod along without understanding, so let us slow down. Angular momentum for a particle moving in a circle is L = mvr — mass times speed times radius. It measures “how much rotation” something carries, in the same way momentum mv measures how much straight-line motion it carries. Its SI unit is J s (joule second), which is also the unit of Planck’s constant — and that is a hint that they belong in the same equation.
Bohr’s claim is that L cannot take just any value. It comes in whole-number packets of h/2π. We write h/2π = ℏ (“h-bar”), so the condition reads L = nℏ.
An everyday analogy that actually helps: think of a stairway rather than a ramp. On a ramp you can stop at any height you like. On a staircase you can only stand on step 1, step 2, step 3 — never at height 2.37 steps. Bohr’s postulate says the electron’s angular momentum is a staircase, and the step size is ℏ = 1.055 × 10−34 J s.
Radius of nth Orbit Derivation Class 12 (Full Working)
This is the derivation that appears most often. Learn the structure, not the algebra: two equations, eliminate v, solve for r.
Equation A — the force condition. Coulomb attraction supplies the centripetal force. For a nucleus of charge +Ze and an electron of charge −e in an orbit of radius rn at speed vn:
(1/4πε0) · Ze²/rn² = mvn²/rn …(A)
Multiply both sides by rn and rearrange:
mvn² = Ze²/(4πε0rn)
Equation B — the quantum condition. Bohr’s second postulate:
mvnrn = nh/2π ⇒ vn = nh/(2πmrn) …(B)
Now eliminate vn. Substitute (B) into the rearranged (A):
m · [nh/(2πmrn)]² = Ze²/(4πε0rn)
n²h²/(4π²mrn²) = Ze²/(4πε0rn)
Multiply both sides by rn² and divide by the right-hand coefficient. The rn² on the left cancels down to leave a single power of rn on the right:
n²h²/(4π²m) = Ze²rn/(4πε0)
and therefore
The headline result: rn ∝ n². Put in the numbers for n = 1, Z = 1 and you get the Bohr radius:
a0 = 5.29 × 10−11 m = 0.529 å
Look at that number and then look back at the scale model. 0.529 å = 0.529 × 10−10 m. Bohr’s theoretical radius, derived from Planck’s constant and the electron charge with no fitting whatsoever, lands squarely on the experimentally known size of a hydrogen atom. That agreement is why this model changed physics.
One step. rn = n²a0 = 9 × 0.529 å = 4.76 å = 4.76 × 10−10 m.
Feel the size: the n = 3 hydrogen atom is nine times wider in radius than the ground state — roughly 1 nm across. Excited atoms are genuinely fat.
Velocity of the Electron in the nth Orbit — and the c/137 Surprise
Once you have rn, the velocity is a single substitution — and this is why doing the radius derivation properly pays off twice.
Start from Bohr’s quantum condition rearranged for v:
vn = nh/(2πmrn)
Substitute rn = n²h²ε0/(πmZe²):
vn = nh · πmZe² / [2πm · n²h²ε0]
Cancel one n, one h, the m, and one π:
The headline result: vn ∝ 1/n. Higher orbits are slower. That should feel right if you think about satellites — a satellite in a high orbit moves more slowly than one skimming the atmosphere, for exactly the same reason: the attractive force out there is weaker, so less speed is needed to keep turning.
Substituting the constants for n = 1, Z = 1:
v1 = 2.19 × 106 m s−1
Now for the genuinely beautiful part. Divide that by the speed of light:
v1/c = 2.19 × 106 / 3.00 × 108 = 7.30 × 10−3 = 1/137
The electron in the ground state of hydrogen moves at almost exactly one hundred and thirty-seventh of the speed of light. This ratio, e²/(2ε0hc), is a pure dimensionless number called the fine-structure constant α, and its value 1/137.036 turns up all over physics. You are not examined on it — but it is a superb thing to know, it makes a memorable one-line remark in a viva, and practically it gives you a lovely mental check: if a Bohr velocity calculation comes out anywhere near c, you have made an arithmetic error.
One step. v4 = v1/4 = 2.19 × 106/4 = 5.47 × 105 m s−1.
As a fraction of c that is 1/(137 × 4) = 1/548 — slower still. The pattern vn = c/(137n) is worth carrying: it turns every velocity question into one division.
Energy of the Electron in the nth Orbit: KE, PE and Total Energy
Total energy is kinetic plus potential, so we build it in three pieces. Do them in this order and you will never mix up the signs.
Piece 1 — kinetic energy. From the force condition (Equation A), mvn² = Ze²/(4πε0rn). Halve it:
KE = ½mvn² = + Ze²/(8πε0rn)
Piece 2 — potential energy. The electrostatic potential energy of a charge −e at distance rn from a charge +Ze, taking zero at infinity, is
PE = (1/4πε0) · (+Ze)(−e)/rn = − Ze²/(4πε0rn)
The negative sign is not decoration. It says the electron is bound — you would have to supply energy to drag it out to infinity. If the idea of a negative potential energy still feels slippery, the underlying convention (zero at infinity, work done by the field) is set out carefully in Coulomb’s law and electric potential energy in the Electric Charges and Fields notes.
Piece 3 — add them.
En = KE + PE = + Ze²/(8πε0rn) − Ze²/(4πε0rn) = − Ze²/(8πε0rn)
Now substitute rn = n²h²ε0/(πmZe²):
En = − Ze² · πmZe² / (8πε0 · n²h²ε0)
Substituting the constants for n = 1, Z = 1 gives E1 = −2.18 × 10−18 J = −13.6 eV. If you would like to see exactly why the electronvolt is such a natural unit here — it is defined by the work a one-volt potential difference does on one electron — that is spelled out in how a potential difference does work on a charge, in the Current Electricity notes.
PE = 2 × TE · KE = −TE = −½PE · TE = −KE
For hydrogen at n = 1: TE = −13.6 eV, KE = +13.6 eV, PE = −27.2 eV. Check: 13.6 − 27.2 = −13.6. These ratios hold for every n and every Z, so you can always get two of the three from one.
Why is the total energy negative? Because the electron is trapped. Negative total energy is the universal signature of a bound system — the same is true of the Moon orbiting the Earth. Zero total energy is the boundary: at E = 0 the electron is just barely free, sitting at infinity with no speed. Positive total energy means a free electron flying away with kinetic energy to spare. That is why the ladder in Figure 1 runs from −13.6 eV up to 0 and then opens into a continuum.
Step 1 — total energy. E3 = −13.6/3² = −13.6/9 = −1.51 eV.
Step 2 — kinetic energy. KE = −TE = +1.51 eV.
Step 3 — potential energy. PE = 2 × TE = −3.02 eV.
Check: KE + PE = 1.51 − 3.02 = −1.51 eV = TE. Consistent.
Total time taken: about fifteen seconds, no calculator. That is what the three relations buy you.
Step 1. E1 = −13.6 eV, E3 = −1.51 eV.
Step 2. Energy required = E3 − E1 = (−1.51) − (−13.6) = +12.09 eV.
Why it works: subtracting a negative number is where marks are lost. Do it as “final minus initial” every single time and let the signs look after themselves. A positive answer means energy absorbed; a negative answer would mean energy emitted. Also note 12.09 eV < 13.6 eV, so this excites the atom but does not ionise it — the electron is lifted to a higher rung, not thrown off the ladder.
Time Period and Orbital Frequency of the Electron
This one is not in the syllabus statement by name, but it is fair game because it is a one-line combination of two results you already have, and it makes a lovely 2-mark question. It also completes the family of four n-dependences we are about to memorise as a unit.
A period is just distance over speed. One lap of a circular orbit is a circumference:
Tn = 2πrn/vn
Now substitute the powers of n. rn ∝ n² and vn ∝ 1/n, so
Tn ∝ n²/(1/n) = n³ and fn = 1/Tn ∝ 1/n³
For n = 1 in hydrogen: T1 = 2π × 5.29 × 10−11 / 2.19 × 106 = 1.52 × 10−16 s, so the orbital frequency is f1 = 6.58 × 1015 revolutions per second.
Sit with that for a moment. The ground-state electron completes six and a half million billion laps every second. In the time it takes you to blink, it has gone round more times than there are seconds in fifty million years. That is why an atom behaves like a smeared-out cloud rather than a tiny solar system with a visible planet in it: on any timescale you can measure, the electron is effectively everywhere on its orbit at once.
Method 1 — from first principles.
r2 = 4 × 0.529 å = 2.117 × 10−10 m and v2 = 2.19 × 106/2 = 1.094 × 106 m s−1.
T2 = 2πr2/v2 = 2π(2.117 × 10−10)/(1.094 × 106) = 1.330 × 10−9/1.094 × 106 = 1.22 × 10−15 s.
Method 2 — from the proportionality (much faster).
T2 = T1 × 2³ = 1.52 × 10−16 × 8 = 1.22 × 10−15 s. Same answer.
Frequency. f2 = 1/T2 = 8.22 × 1014 Hz.
Why it works: Method 2 is what you should use in an exam. Once you trust T ∝ n³, every period question collapses to “T1 times n cubed” — one multiplication instead of four.
The 2-Spine Trick: One Memory Device for r ∝ n², v ∝ 1/n, E ∝ 1/n², T ∝ n³
Four results, four different powers of n, and about a third of all marks lost in this chapter come from swapping two of them over in the exam hall. So here is the device I want you to use instead of rote memorising a list of four. I call it the 2-Spine Trick, and the whole point is that you memorise exactly one power and generate the other three.
Then generate the rest with three one-line rules, each of which is a physical statement, not a trick:
1. Speed borrows one from the spine. Because mvr = nh/2π, we have v ∝ n/r ∝ n1/n2. Power of v = 1 − 2 = −1.
2. Energy flips the spine. Because E = −ke²/2r, energy depends on 1/r. Power of E = −(power of r) = −2.
3. A lap is distance over speed. Because T = 2πr/v, power of T = (power of r) − (power of v) = 2 − (−1) = +3.
Order of use: R → V → E → T. Say it as “RaVET rides one spine”: Radius, Velocity, Energy, Time — two, minus one, minus two, plus three.
Why is this better than memorising “2, −1, −2, 3”? Because a memorised list can be mis-ordered and you will never know. A generated list self-checks. If you write down v ∝ n by mistake, rule 1 immediately contradicts you: it demands 1 minus 2, which cannot be positive. Every one of the three rules is one honest line of algebra you can redo in the margin in five seconds.
Now the physical intuition for each power, because understanding beats remembering:
| Quantity | n-dependence | Why physically |
|---|---|---|
| Radius rn | ∝ n² | Angular momentum grows as n while the Coulomb grip weakens with distance. To carry n times the angular momentum, the electron must move out disproportionately far — the orbit inflates as the square. |
| Velocity vn | ∝ 1/n | Same reason a distant satellite is slow. The pull out there is weak, so only a small speed is needed to keep the electron turning. Outer orbits are lazy orbits. |
| Total energy En | ∝ −1/n² | Energy is set by how deep in the electrostatic well you sit, and depth goes as 1/r. Since r ∝ n², the well gets shallow fast — which is exactly why the ladder rungs in Figure 1 crowd together at the top. |
| Time period Tn | ∝ n³ | A bigger loop travelled more slowly: the lap time gets hit twice. This is Kepler’s third law in disguise — T² ∝ r³ gives T ∝ (n²)3/2 = n³. Same maths, different century. |
That last row is a genuinely satisfying check. Bohr’s quantised electron obeys Kepler’s third law exactly, because both are consequences of an inverse-square attraction. If you ever forget the cube, recover it from Kepler.
Run RaVET.
(a) r3/r1 = 3² = 9
(b) v3/v1 = 1/3 = 0.333
(c) |E3|/|E1| = 1/3² = 1/9
(d) T3/T1 = 3³ = 27
Cross-check with the actual numbers: r3 = 4.76 å vs r1 = 0.529 å → 9.0 ✓. |E3| = 1.51 eV vs 13.6 eV → 1/9.0 ✓. T3 = 4.10 × 10−15 s vs 1.52 × 10−16 s → 27.0 ✓. Every one lands.
Why it works: ratio questions never need the constants. Spot the word “ratio” or “compare” and you should not be reaching for your calculator at all.
Hydrogen Line Spectra (Qualitative Treatment Only) and the Energy-Level Diagram
Read the syllabus wording carefully: hydrogen line spectra, qualitative treatment only. That is genuinely good news. CBSE wants you to understand and describe the origin of the spectral series, read an energy-level diagram, and use hν = Ei − Ef to work out a wavelength. You are not required to derive the Rydberg formula from scratch or produce fine-structure calculations.
Here is how the barcode is born. Take a glass tube of hydrogen gas at low pressure and pass an electric discharge through it. The discharge kicks electrons up to higher levels — they get excited. An excited electron does not stay up there; within roughly 10−8 s it drops back down, and by Bohr’s third postulate each drop emits one photon of energy exactly equal to the gap it fell through.
Because the energy levels are a fixed ladder, the set of possible gaps is fixed, so the set of possible photon energies is fixed, so the set of emitted wavelengths is fixed. Different atoms in the gas take different routes down, so all the possible lines appear at once — and what you see through a spectroscope is a set of sharp coloured lines on a dark background. A line emission spectrum. The barcode is a direct photograph of the energy ladder.
From hν = Ei − Ef with En = −13.6/n² eV, and using c = νλ:
In practice, for CBSE numericals, the single most useful form is the energy version with a conversion baked in:
ΔE (eV) = 13.6 (1/nf² − 1/ni²) and λ (nm) = 1240 / ΔE (eV)
Now look back at Figure 1 with that in hand. The rungs are drawn at their true energies, and the two things that jump out are exactly the two things examiners test. First, the n = 1 rung is enormously far below all the others — which is why Lyman transitions have the biggest energy gaps and therefore the shortest wavelengths, right out in the ultraviolet. Second, the rungs above n = 3 crowd together brutally — which is why the higher series lie in the infrared and why the lines within any series bunch up as you approach the series limit.
Step 1 — the two energies. E4 = −13.6/16 = −0.85 eV; E2 = −13.6/4 = −3.40 eV.
Step 2 — the gap. ΔE = −0.85 − (−3.40) = 2.55 eV.
Step 3 — the wavelength. λ = 1240/2.55 = 486 nm.
Sanity check: 486 nm is blue-green, and this is the second Balmer line, H-β, which is indeed seen as a blue-green line in a hydrogen discharge tube. Whenever a Balmer answer comes out between about 365 nm and 660 nm you are almost certainly right; anything outside that window means you have used the wrong nf.
Step 1 — what “shortest” means. Shortest wavelength = biggest energy gap = the electron falls from as high as possible, i.e. from ni = ∞. Balmer means nf = 2.
Step 2. ΔE = E∞ − E2 = 0 − (−3.40) = 3.40 eV.
Step 3. λ = 1240/3.40 = 365 nm (computing with 13.6057 eV gives 364.5 nm).
Why it works: a series limit is always the transition from infinity, so the energy gap is simply |Enf|. This makes limits the easiest questions in the chapter: the Lyman limit is 1240/13.6 ≈ 91 nm, the Balmer limit is 1240/3.40 ≈ 365 nm, the Paschen limit is 1240/1.51 ≈ 820 nm. Three divisions, three marks.
Hydrogen Spectrum Series Formula List: Lyman, Balmer, Paschen, Brackett, Pfund
Five named series. The name tells you the landing level and nothing else. That is the single most important sentence in this section, so read it again.
| Series | Lands on nf | Comes from ni | Region | First line (longest λ) | Series limit (shortest λ) |
|---|---|---|---|---|---|
| Lyman | 1 | 2, 3, 4, … | Ultraviolet | 121.5 nm (2→1) | 91.1 nm |
| Balmer | 2 | 3, 4, 5, … | Visible | 656.1 nm (3→2, red H-α) | 364.5 nm |
| Paschen | 3 | 4, 5, 6, … | Infrared | 1874.6 nm (4→3) | 820.1 nm |
| Brackett | 4 | 5, 6, 7, … | Far infrared | 4050 nm (5→4) | 1458 nm |
| Pfund | 5 | 6, 7, 8, … | Far infrared | 7456 nm (6→5) | 2278 nm |
How to remember the order. The landing levels run 1, 2, 3, 4, 5 in the order Lyman, Balmer, Paschen, Brackett, Pfund. Use the sentence “Little Boys Play Badminton Properly” — five words, five series, in landing-level order 1 to 5. The two Bs and the two Ps are the only genuine hazard, and the sentence fixes their order for you.
The wavelengths in the table are computed from En = −13.6057/n² eV using λ = 1239.84/ΔE. You will meet slightly different values in different books — 91.2 nm instead of 91.1 nm for the Lyman limit, 364.6 nm instead of 364.5 nm for Balmer — purely because some books round 13.6057 eV to 13.60 eV before dividing. Both are accepted. Quote your answers to three significant figures and you will never be marked down on this.
- The longest wavelength (lowest energy, first line, the “α” line) always comes from the level immediately above the landing level: nf+1 → nf.
- The shortest wavelength (series limit) always comes from ni = ∞, and equals 1240/|Enf| nm. Lines pile up ever closer together as you approach it.
Which Series Does This Line Belong To? A Four-Step Decision Procedure
Here is the promise from the introduction, delivered. A question gives you a wavelength — say 1094 nm — and asks which series it belongs to and which transition produced it. Under exam pressure, students guess. You are going to run a procedure instead. Four steps, and it never fails.
- Convert to energy. ΔE (eV) = 1240 / λ (nm). One division. Do it first, always — energies are far easier to reason about than wavelengths.
- Bracket it against the limits. Compare ΔE with the four landing-level depths: 13.61 eV (n=1), 3.40 eV (n=2), 1.51 eV (n=3), 0.85 eV (n=4). The series is the largest landing depth that is still bigger than your ΔE. (A line of a series can never exceed its own series limit energy.)
- Find ni. With nf now known, solve 1/ni² = 1/nf² − ΔE/13.6, then take the square root. It must come out to (very nearly) a whole number.
- Verify. Recompute λ from your ni and nf. If it matches the given wavelength to within rounding, you are done. If not, you picked the wrong nf in step 2 — move one series along and repeat.
Step 2 is the clever one, so let us make it concrete. Suppose ΔE = 1.13 eV. Is it Lyman? A Lyman line needs at least 10.2 eV (the smallest possible Lyman gap), so no. Balmer? The smallest Balmer gap is 1.89 eV, and 1.13 < 1.89, so no. Paschen? The smallest Paschen gap is 0.66 eV and the Paschen limit is 1.51 eV, and 0.66 < 1.13 < 1.51, so yes. It is a Paschen line. You have identified the series with two comparisons and no algebra.
| Landing level | Smallest gap in the series (first line) | Series limit (largest gap) | So ΔE must lie in |
|---|---|---|---|
| nf = 1 (Lyman) | 10.20 eV | 13.61 eV | 10.20 – 13.61 eV |
| nf = 2 (Balmer) | 1.89 eV | 3.40 eV | 1.89 – 3.40 eV |
| nf = 3 (Paschen) | 0.66 eV | 1.51 eV | 0.66 – 1.51 eV |
| nf = 4 (Brackett) | 0.31 eV | 0.85 eV | 0.31 – 0.85 eV |
| nf = 5 (Pfund) | 0.17 eV | 0.54 eV | 0.17 – 0.54 eV |
Notice that these energy windows do not overlap. That is the deep reason the procedure is reliable: a hydrogen photon energy determines its series uniquely. Copy this small table into your formula sheet; it converts a whole class of questions into a lookup.
Step 1 — convert. ΔE = 1240/1094 = 1.133 eV.
Step 2 — bracket. 1.133 eV lies in the window 0.66 – 1.51 eV → Paschen series, nf = 3.
Step 3 — find ni.
1/ni² = 1/9 − 1.133/13.6057 = 0.11111 − 0.08327 = 0.02784
ni² = 1/0.02784 = 35.9 → ni = √35.9 = 5.99 ≈ 6.
Step 4 — verify. ΔE(6→3) = 13.6057(1/9 − 1/36) = 13.6057 × 0.08333 = 1.134 eV; λ = 1239.84/1.134 = 1093.5 nm. Matches 1094 nm.
Answer: the 6 → 3 transition, third line of the Paschen series, in the infrared. Total time under exam conditions: about ninety seconds, and every step is checkable.
Emission vs Absorption Spectra, Excitation Energy and Ionisation Energy
Two spectra, two words, and one very common muddle. Sort it out with the direction of the jump.
| Emission spectrum | Absorption spectrum | |
|---|---|---|
| What the electron does | Jumps down: ni → nf, ni > nf | Jumps up: nf → ni, absorbing a photon |
| What the photon does | A photon is created | A photon is destroyed |
| How you produce it | Pass a discharge through hydrogen gas, then look at the light it gives out | Shine white light through cool hydrogen gas, then look at what comes out |
| What you see | Bright coloured lines on a dark background | Dark lines on a continuous bright background |
| Where the lines sit | At wavelengths λ = 1240/ΔE | At exactly the same wavelengths |
That last row is the punchline and it is examinable: the emission and absorption line positions of an element are identical, because both are governed by the same set of energy gaps. Hydrogen absorbs precisely the colours it emits. This is how astronomers identify which elements are present in a star they will never visit — they read the dark lines in the star’s spectrum like a fingerprint. If you enjoyed the way a prism separates white light into its components in Class 10 Light: Reflection and Refraction, a spectroscope is that same dispersion used as a measuring instrument.
Ionisation energy is the energy needed to remove the electron completely, i.e. take it to n = ∞ where E = 0. From the ground state of hydrogen that is 0 − (−13.6) = 13.6 eV.
Ionisation potential is the same number quoted in volts: 13.6 V. Energy in eV and potential in volts are numerically equal for a single electron, which is exactly why the electronvolt is used here.
An important subtlety about ionisation energy: it depends on where you start. From the ground state it is 13.6 eV, but from the n = 2 state it is only 0 − (−3.40) = 3.40 eV, and from n = 3 only 1.51 eV. An excited atom is far easier to ionise. Always check which level the question starts from.
One more result that questions love, and it is pure counting rather than physics. If a hydrogen atom is excited to level n, how many different spectral lines can appear as it cascades back down? Every possible pair of levels from 1 up to n gives one line, so the count is the number of ways of choosing 2 levels from n:
Number of spectral lines = n(n − 1)/2
So n = 2 gives 1 line, n = 3 gives 3 lines, n = 4 gives 6 lines, n = 5 gives 10 lines, and n = 6 gives 15 lines.
(a) What was the energy of the absorbed photon, and its wavelength?
(b) How many different spectral lines can the atom emit as it returns to the ground state?
(c) Which of those lines has the longest wavelength, and what is it?
(d) Which has the shortest, and what is it?
(a) ΔE = E4 − E1 = −0.85 − (−13.61) = 12.76 eV. λ = 1239.84/12.755 = 97.2 nm (ultraviolet).
(b) n(n−1)/2 = 4 × 3/2 = 6 lines. They are 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1.
(c) Longest λ = smallest ΔE. Compare the six gaps: 4→3 is 0.661 eV, which is the smallest (every other gap is larger). λ = 1239.84/0.6614 = 1875 nm — the first Paschen line, in the infrared.
(d) Shortest λ = largest ΔE = 4→1 = 12.755 eV. λ = 97.2 nm — a Lyman line. Note it is the same wavelength as the photon absorbed in part (a), which is a satisfying consistency check: the atom can give back in one go exactly what it took in one go.
Why it works: parts (c) and (d) never need all six wavelengths computed. Longest λ is always the smallest step, which is always the top step (n → n−1). Shortest λ is always the whole drop (n → 1). Two calculations, not six.
Limitations of Bohr’s Model
Bohr’s model was a triumph for hydrogen and a dead end almost everywhere else. Examiners ask for its limitations regularly, usually for 2 or 3 marks, so learn a clean list. Group them as what it cannot handle and why it was philosophically unsatisfying.
- It works only for single-electron (hydrogen-like) atoms. Hydrogen, He+, Li2+ — fine. Neutral helium onwards — it fails, because with two or more electrons there are electron–electron repulsions that the model has no way to include.
- It cannot explain the relative intensities of spectral lines. The model tells you which wavelengths appear, but not why some lines are bright and others faint. It has no notion of transition probability.
- It cannot explain the fine structure of spectral lines. Under high resolution, single lines turn out to be several very closely spaced lines. Bohr’s single quantum number n cannot produce that splitting.
- It cannot explain the splitting of lines in magnetic or electric fields (the Zeeman and Stark effects). A model with circular orbits and one quantum number has no way to respond to field direction.
- It says nothing about chemical bonding or molecular spectra. Why hydrogen forms H2, why water is bent — all outside its reach.
- It is a hybrid, and an uncomfortable one. It uses classical mechanics for the orbit and then arbitrarily forbids the classical radiation that mechanics demands. It offers no reason why angular momentum should be quantised — the postulate is imposed, not explained.
- It contradicts the uncertainty principle. A definite orbit means simultaneously knowing position and momentum precisely, which quantum mechanics forbids. Modern physics replaces the orbit with an orbital — a probability cloud.
None of this makes Bohr’s work a failure. It was the first model to get a spectrum right from theory, it introduced the quantum number, and it gave the ionisation energy of hydrogen correct to three significant figures from first principles. It was replaced, in the 1920s, by full quantum mechanics — which keeps every one of Bohr’s energy levels for hydrogen and explains where they come from. That is the right way to think about it: not wrong, but incomplete. If you want to see where this thread picks up next, the wave description of matter is developed in the neighbouring chapter of the full Class 12 Physics chapter list.
Reference Only — Adjacent Topics Not in Your 2026-27 CBSE Exam
1. De Broglie’s explanation of Bohr’s quantisation condition. This is the big one, and it has been removed from the 2026-27 syllabus for this chapter — verified against the official curriculum document, where it does not appear at all. The idea is elegant, so here it is in three lines. Treat the electron as a de Broglie wave of wavelength λ = h/mv. For the wave to be a stable standing wave around a circular orbit, the circumference must contain a whole number of wavelengths: 2πr = nλ = nh/mv. Rearranging gives mvr = nh/2π — exactly Bohr’s second postulate. The mysterious integer n becomes simply “the number of wave crests that fit around the loop”. Beautiful, but not examinable for you.
2. Hydrogen-like ions with Z > 1. The CBSE line specifies the hydrogen atom. Every formula above generalises by keeping Z: rn ∝ n²/Z, vn ∝ Z/n, En = −13.6Z²/n² eV. So He+ has a ground state at −54.4 eV and Li2+ at −122.4 eV. Competitive exams use this constantly; your board paper will stay at Z = 1, but knowing the Z-scaling costs you nothing and makes the derivation feel less like a special case.
3. The Rydberg constant derived from Bohr’s theory. Combining the energy expression with hν = ΔE gives R = me4/(8ε0²h³c), which evaluates to 1.0974 × 107 m−1 — matching the spectroscopically measured value. This derivation is not required under “qualitative treatment only”; you simply use R as a given constant.
4. Radioactivity, α/β/γ decay, half-life and the radioactive decay law. These are all absent from the 2026-27 syllabus — verified against the official document, which contains zero occurrences of them. Students frequently trip over this because the α-source in the Geiger–Marsden experiment is a radioactive source, and older notes bolt a whole radioactivity section on to Chapter 12 or 13. For your board exam you need to know only that α-particles are helium nuclei with charge +2e and mass about 4 u. You do not need decay laws, half-life, mean life, or the properties of β and γ radiation.
5. Fine structure, Zeeman and Stark effects, and the four quantum numbers. Mentioned above only as limitations of Bohr’s model, which is examinable. You are not expected to explain or calculate them.
Practice Worksheet — Atoms Class 12 Physics Important Questions
Ten original questions, ordered roughly from easiest to hardest. Work each one on paper before you tap Show Answer — reading a solution feels like learning, but writing one is learning. Constants: a0 = 0.529 å, E1 = −13.6 eV, v1 = 2.19 × 106 m s−1, T1 = 1.52 × 10−16 s, 1/4πε0 = 8.99 × 109 N m² C−2, e = 1.602 × 10−19 C, λ(nm) = 1240/ΔE(eV). Give answers to 3 significant figures.
Q1. Find the radius of the n = 5 orbit of a hydrogen atom, and state how many times larger it is than the ground-state radius.
It is n² = 25 times the ground-state radius. (No calculator needed for the second part — that is the spine of the 2-Spine Trick doing its job.)
Q2. Find (a) the speed of the electron in the n = 3 orbit of hydrogen and (b) the time it takes to complete one revolution in that orbit.
(b) Tn = T1n³ = 1.52 × 10−16 × 27 = 4.10 × 10−15 s.
Check by first principles: r3 = 9 × 0.529 å = 4.76 × 10−10 m, so T = 2πr/v = 2π(4.76 × 10−10)/(7.30 × 105) = 2.99 × 10−9/7.30 × 105 = 4.10 × 10−15 s. Agrees.
Q3. A hydrogen atom is in the n = 4 state. Write down its total energy, kinetic energy and potential energy in eV, and verify that they are consistent.
Kinetic energy: KE = −TE = +0.850 eV.
Potential energy: PE = 2 × TE = −1.70 eV.
Consistency check: KE + PE = 0.850 − 1.70 = −0.850 eV = TE ✓. Also PE/KE = −2 ✓.
Q4. Calculate the wavelength of the radiation emitted when a hydrogen atom makes a transition from n = 5 to n = 2. Which series does it belong to, and what colour is it?
ΔE = −0.544 − (−3.40) = 2.86 eV.
λ = 1240/2.856 = 434 nm.
Since it lands on n = 2 it is the Balmer series; specifically the third Balmer line, H-γ. 434 nm is in the violet part of the visible spectrum.
Q5. Find the series limit (shortest wavelength) of the Paschen series, and explain in one sentence what a series limit physically represents.
ΔE = 0 − (−13.6/9) = |E3| = 1.51 eV.
λ = 1240/1.511 = 820 nm (computing with 13.6057 eV gives 820.1 nm).
Physical meaning: the series limit is the photon emitted when a completely free electron (at n = ∞, zero energy) is captured directly into the n = 3 level — the largest possible energy drop that still lands on n = 3, and hence the shortest possible wavelength in that series.
Q6. A 7.7 MeV α-particle is fired head-on at a gold nucleus (Z = 79). Calculate the distance of closest approach.
r0 = (1/4πε0)(2Ze²/K) = 8.99 × 109 × 2 × 79 × (1.602 × 10−19)² / 1.234 × 10−12
Numerator = 8.99 × 109 × 158 × 2.567 × 10−38 = 3.646 × 10−26
r0 = 3.646 × 10−26/1.234 × 10−12 = 2.95 × 10−14 m = 29.5 fm.
Shortcut check: r0 ∝ 1/K, so from the 5.5 MeV answer of 41.4 fm we get 41.4 × (5.5/7.7) = 29.6 fm. Agrees.
Q7. For the same gold target and a 5.5 MeV α-particle, find the impact parameter that produces a scattering angle of 60°. Compare it with the value for 90° and explain the trend.
The 90° case (cot 45° = 1) gives b90 = 20.7 fm, so
b60 = 20.7 × 1.732 = 35.8 fm.
Trend and reason: the impact parameter for 60° is larger than for 90°. A larger impact parameter means the α-particle passes further from the nucleus, feels a weaker Coulomb repulsion, and is therefore deflected through a smaller angle. Impact parameter and scattering angle always move in opposite directions.
Q8. A line in the hydrogen spectrum has wavelength 1875 nm. Identify the series and the transition, using the four-step decision procedure.
Step 2 — bracket. 0.661 eV lies in the Paschen window 0.66 – 1.51 eV, so nf = 3, Paschen series. (It is also infrared, which rules out Lyman and Balmer immediately.)
Step 3 — find ni. 1/ni² = 1/9 − 0.661/13.6057 = 0.11111 − 0.04858 = 0.06253; ni² = 15.99 → ni = 4.
Step 4 — verify. ΔE(4→3) = 13.6057(1/9 − 1/16) = 13.6057 × 0.048611 = 0.6614 eV; λ = 1239.84/0.6614 = 1874.6 nm. Matches 1875 nm.
Answer: the 4 → 3 transition, the first (longest-wavelength) line of the Paschen series.
Q9. A hydrogen atom is excited to the n = 3 level. (a) How many spectral lines can it emit on returning to the ground state? (b) List each transition with its wavelength. (c) Which of them are visible?
(b) Energies: E1 = −13.61 eV, E2 = −3.401 eV, E3 = −1.512 eV.
3 → 2: ΔE = 1.890 eV → λ = 1239.84/1.890 = 656 nm (Balmer H-α, red)
2 → 1: ΔE = 10.20 eV → λ = 1239.84/10.204 = 122 nm (Lyman-α, UV)
3 → 1: ΔE = 12.09 eV → λ = 1239.84/12.094 = 103 nm (Lyman, UV)
(c) Only the 656 nm line is visible — it is the familiar red H-α. The other two are ultraviolet, because every transition ending on n = 1 has a gap of at least 10.2 eV, far too energetic for visible light.
Q10. (a) How much energy is needed to ionise a hydrogen atom that is already in the n = 3 state? (b) A photon of energy 2.00 eV strikes such an atom — can it ionise it? (c) A photon of energy 12.00 eV strikes a hydrogen atom in its ground state. Can it be absorbed? Justify your answers.
Energy needed = 0 − E3 = 0 − (−1.51) = 1.51 eV.
(b) Yes. 2.00 eV exceeds the 1.51 eV required, so the atom is ionised and the freed electron carries away the surplus as kinetic energy: 2.00 − 1.51 = 0.49 eV. (Ionisation is not a resonance condition — any energy above the threshold works, because the free electron can take any energy at all.)
(c) No. From the ground state the allowed absorptions are exactly the level gaps: 1→2 needs 10.20 eV, 1→3 needs 12.09 eV, 1→4 needs 12.75 eV, and full ionisation needs 13.61 eV. A 12.00 eV photon matches none of these and is below the ionisation energy, so it cannot be absorbed and passes straight through.
The lesson: bound-to-bound transitions require an exact energy match; bound-to-free (ionising) transitions require only a minimum. Mixing those two rules up is one of the most common errors in this chapter.